hdu 5793 A Boring Question(2016 Multi-University Training Contest 6——快速幂取模)

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最新推荐文章于 2017-07-16 09:41:00 发布

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分类专栏： ACM 快速幂取模 2016 Multi-University Training Contest I believe i can

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本文针对HDU 5793竞赛题ABoringQuestion进行详细解析，介绍了解题思路及核心代码实现，通过快速幂运算解决特定数学公式计算问题。

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5793

A Boring Question

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 453 Accepted Submission(s): 249

Problem Description

There are an equation.

We define that

. And

while

.
You have to get the answer for each

and

that given to you.
For example,if

,
When

;
When

.
So the answer is 4.

Input

The first line of the input contains the only integer

Then

lines follow,the i-th line contains two integers

Output

For each

and

,output the answer in a single line.

Sample Input

Sample Output

  
   3
13

Author

UESTC

Source

2016 Multi-University Training Contest 6

题目大意：求给出公式的值。

解题思路：

（其实也可以打表找到这个规律）

详见代码。

#include <iostream>
#include <cstdio>

using namespace std;

#define ll long long
#define mod 1000000007

ll fun(ll a,ll b)
{
    ll s=1;
    while (b)
    {
        if (b%2==1)
            s=s*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return s%mod;
}

int main()
{
    ll t,n,m;
    scanf("%lld",&t);
    while (t--)
    {
        scanf("%lld%lld",&n,&m);
        ll ans=((fun(m,n+1)-1)*fun(m-1,mod-2)%mod+mod)%mod;
        printf ("%lld\n",ans);
    }
}